CH 9: The Mole

Renee Y. BeckerCHM 1025

Valencia Community College

1

2

• Avogadro’s number (symbol N) is the number of atoms in 12.01 grams of carbon.

• Its numerical value is 6.02 × 1023.

• Therefore, a 12.01 g sample of carbon contains 6.02 × 1023 carbon atoms.

Avogadro’s Number

3

• The mole (mol) is a unit of measure for an amount of a chemical substance.

• A mole is Avogadro’s number of particles, which is 6.02 × 1023 particles.

1 mol = Avogadro’s number = 6.02 × 1023 units

• We can use the mole relationship to convert between the number of particles and the mass of a substance.

The Mole

4

• The volume occupied by one mole of softballs would be about the size of the Earth.

• One mole of Olympic shotput balls has about the same mass as the Earth.

Analogies for Avogadro’s Number

5

One Mole of Several Substances

C12H22O11 H2Omercury

sulfur

NaClcopper

lead

K2Cr2O7

6

• How many sodium atoms are in 0.120 mol Na?

Example 1

7

• How many moles of potassium are in 1.25 × 1021 atoms K?

Example 2

8

• The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.

• The atomic mass of iron is 55.85 amu.

• Therefore, the molar mass of iron is 55.85 g/mol.

• Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol.

Molar Mass

9

• The molar mass of a substance is the sum of the molar masses of each element.

• What is the molar mass of magnesium nitrate, Mg(NO3)2?

• The sum of the atomic masses is:

24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) =

24.31 + 2(62.01) = 148.33 amu

• The molar mass for Mg(NO3)2 is 148.33 g/mol.

Calculating Molar Mass

Example 3

Calc. the molar mass:

1. Fe2O3

2. C6H8O7

3. C16H18N2O4

10

11

• Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance.

6.02 × 1023 particles = 1 mol = molar mass

• If we want to convert particles to mass, we must first convert particles to moles and then we can convert moles to mass.

Mole Calculations

12

• What is the mass of 1.33 moles of titanium, Ti?

Example 4

13

• What is the mass of 2.55 × 1023 atoms of lead?

Example 5

14

• How many O2 molecules are present in 0.470 g of oxygen gas?

Example 6

15

• What is the mass of a single molecule of sulfur dioxide? The molar mass of SO2 is 64.07 g/mol.

Example 7

16

• At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.

• The volume occupied by 1 mole of gas (22.4 L) is called the molar volume.

• Standard temperature and pressure are 0C and 1 atm.

Molar Volume

17

• We now have a new unit factor equation:1 mole gas = 6.02 × 1023 molecules gas = 22.4 L gas

Molar Volume of Gases

18

• The density of gases is much less than that of liquids.

• We can calculate the density of any gas at STP easily.

• The formula for gas density at STP is:

= density, g/Lmolar mass in grams/molmolar volume in liters/mol

Gas Density

19

• What is the density of ammonia gas, NH3, at STP?

Example 8

20

• We can also use molar volume to calculate the molar mass of an unknown gas.

• 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass?

• We want g/mol; we have g/L.

Example 9

21

• We now have three interpretations for the mole:

–1 mol = 6.02 × 1023 particles

–1 mol = molar mass

–1 mol = 22.4 L at STP for a gas

• This gives us 3 unit factors to use to convert between moles, particles, mass, and volume.

Mole Unit Factors

22

• A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?

Example 10

23

• What is the mass of 3.36 L of ozone gas, O3, at STP?

Example 11

24

• How many molecules of hydrogen gas, H2, occupy 0.500 L at STP?

Example 12

25

• The percent composition of a compound lists the mass percent of each element.

• For example, the percent composition of water, H2O is:– 11% hydrogen and 89% oxygen

• All water contains 11% hydrogen and 89% oxygen by mass.

Percent Composition

26

• There are a few steps to calculating the percent composition of a compound. Let’s practice using H2O.

– Assume you have 1 mole of the compound.

– One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen.

– 2(1.01 g H) + 1(16.00 g O) = molar mass H2O

– 2.02 g H + 16.00 g O = 18.02 g H2O

Calculating Percent Composition

27

• Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water:

2.02 g H18.02 g H2O

× 100% = 11.2% H

16.00 g O18.02 g H2O

× 100% = 88.79% O

Calculating Percent Composition

% H =

% O =

28

• TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3.

Example 13

29

Percent Composition of TNT

30

• The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule.

• The molecular formula of benzene is C6H6.

– The empirical formula of benzene is CH.

• The molecular formula of octane is C8H18.

– The empirical formula of octane is C4H9.

Empirical Formulas

31

• We can calculate the empirical formula of a compound from its composition data.

• We can determine the mole ratio of each element from the mass to determine the formula of radium oxide, Ra?

O?.

• A 1.640 g sample of radium metal was heated to produce 1.755 g of radium oxide. What is the empirical formula?

• We have 1.640 g Ra and 1.755-1.640 = 0.115 g O.

Calculating Empirical Formulas

32

• The molar mass of radium is 226.03 g/mol, and the molar mass of oxygen is 16.00 g/mol.

• We get Ra0.00726O0.00719. Simplify the mole ratio by dividing by the smallest number.

• We get Ra1.01O1.00 = RaO is the empirical formula.

Calculating Empirical Formulas

1 mol Ra226.03 g Ra

1.640 g Ra × = 0.00726 mol Ra

1 mol O16.00 g O

0.115 g O × = 0.00719 mol O

Example 14

• Determine the empirical formula of sodium sulfate given the following data

A 3.00 g sample of NaxSyOz is comprised of 0.9722 g of sodium, 0.678 g of sulfur and 1.35 g of oxygen

33

34

• We can also use percent composition data to calculate empirical formulas.

• Assume that you have 100 grams of sample.

• Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula?

• If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen.

Empirical Formulas from Percent Composition

35

• Calculate the moles of each element:

1 mol C12.01 g C

92.2 g C × = 7.68 mol C

1 mol H1.01 g H

7.83 g H × = 7.75 mol H

• The ratio of elements in acetylene is C7.68H7.75. Divide by the smallest number to get the formula:

7.687.68C = C1.00H1.01 = CH7.75

7.68H

Empirical Formula for Acetylene

Example 15

• Find the empirical formula of Iron(III) chloride from the following data

• In Iron(III) chloride– Fe 34.43 % by mass–Cl 65.57 % by mass

36

37

• The empirical formula for acetylene is CH. This represents the ratio of C to H atoms on acetylene.

• The actual molecular formula is some multiple of the empirical formula, (CH)n.

• Acetylene has a molar mass of 26 g/mol. Find n to find the molecular formula:

=CH

(CH)n 26 g/mol13 g/mol

n = 2 and the molecular formula is C2H2.

Molecular Formulas

Example 16

• Find the empirical and molecular formula for hydrazine, NxHy

• The molar mass of hydrazine is 32.046 g/mol

• In hydrazine– N is 87.42% by mass– H is 12.58 % by mass

38

39

• We can use the following flow chart for mole calculations:

Chapter Summary